Question 287: Waves

Question

Two sound waves having same amplitude ' $A$ ' and angular frequency ' $\omega$ ' but having a phase difference of $\left(\frac{\pi}{2}\right)^c$ are superimposed then the maximum amplitude of the resultant wave is

Accepted Answer

Let's consider the two waves given by
$$ y_1 = A \sin(\omega t) $$
and
$$ y_2 = A \sin\left(\omega t + \frac{\pi}{2}ight). $$
When adding two sine waves of equal amplitude $ A $ with a phase difference $ \delta $, we can use the identity:
$$ \sin 	heta + \sin(	heta + \delta) = 2 \sin\left(	heta + \frac{\delta}{2}ight) \cos\left(\frac{\delta}{2}ight). $$
Here, $ 	heta = \omega t $ and $ \delta = \frac{\pi}{2} $. Plugging in the phase difference gives:
$$ y_1 + y_2 = 2A \sin\left(\omega t + \frac{\pi}{4}ight) \cos\left(\frac{\pi}{4}ight). $$
Since
$$ \cos\left(\frac{\pi}{4}ight) = \frac{\sqrt{2}}{2}, $$
the expression simplifies to:
$$ y_1 + y_2 = 2A \cdot \frac{\sqrt{2}}{2} \sin\left(\omega t + \frac{\pi}{4}ight) = A\sqrt{2}\, \sin\left(\omega t + \frac{\pi}{4}ight). $$
This shows that the maximum amplitude of the resultant wave is:
$$ A\sqrt{2}. $$
Thus, the correct answer is:
Option C: $ \sqrt{2}\, A $.

Answer

$\frac{\mathrm{A}}{\sqrt{2}}$

Answer

$\frac{\mathrm{A}}{2}$

Answer

$\sqrt{2} \mathrm{~A}$

Answer

$\mathrm{2 A}$

Waves

Two sound waves having same amplitude ' A ' and angular frequency ' \omega ' but having a phase difference of \left(\frac{\pi}{2}\right)^c are superimposed then the maximum amplitude of the resultant wave is

Choose the correct answer: